# Question fd6e0

Oct 3, 2016

Approx. $30 \cdot m g$ ${O}_{2}$ gas are required.

#### Explanation:

We need a stoichiometrically balanced equation:

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right)$

Is this balanced? Why?

The equation tells us unequivocally that 1 mole of methane gas requires 2 moles of dioxygen gas for complete combustion.

$\text{Moles of methane}$ $=$ $\frac{7.90 \times {10}^{-} 3 \cdot g}{16.04 \cdot g \cdot m o {l}^{-} 1}$ $= 4.93 \times {10}^{-} 4 \cdot m o l$.

And thus 2 equiv of dioxygen gas are required (I use the term $\text{dioxygen}$ to avoid confusion with regard to the number of oxygen atoms I need) per equiv of methane gas.

$\text{Moles of dioxygen}$ $=$ 4.93xx10^-4*cancel(mol)xx2xx32.00*g*cancel(mol^-1)=??g#